A certain metal crystallizes in a bcc lattice | vedclass

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(B) In a $bcc$ lattice, the nearest neighbours are at a distance of $d_1 = \frac{\sqrt{3} a}{2}$, where $a$ is the edge length of the unit cell.Given

Vedclass · Accepted Answer

$500$

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(B) In a $bcc$ lattice, the nearest neighbours are at a distance of $d_1 = \frac{\sqrt{3} a}{2}$, where $a$ is the edge length of the unit cell.Given $d_1 = 433 	ext{ pm}$, we have $\frac{\sqrt{3} a}{2} = 433 	ext{ pm}$.Therefore, the edge length $a = \frac{2 	imes 433}{\sqrt{3}} = \frac{866}{1.732} \approx 500 	ext{ pm}$.The next nearest neighbours in a $bcc$ lattice are located at the corners of the unit cell along the edge, so the distance $d_2 = a$.Thus, the distance between the next nearest neighbours is $500 	ext{ pm}$.

$A$ certain metal crystallizes in a $bcc$ lattice. The distance between the nearest neighbours is $433 \text{ pm}$. The distance between the next nearest neighbours is :- .............. $\text{pm}$

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